Algebra precalculus

     
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I can"t figure out how to lớn simplify $sin2 heta cos2 heta$ khổng lồ be $frac12 sin4 heta $. I assume double angle identities are involved, but I can"t get it & would lượt thích some help.

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We know the double angle formula for sine is $sin(2x) = 2sin(x)cos(x)$.

For convenience, let $x = 2 heta$. Then $4 heta$ can be written as$$4 heta = 2(2 heta) = 2x.$$It then follows that$$frac12sin(4 heta) = frac12sin(2x) = frac12 cdot 2sin(x)cos(x) = sin(x)cos(x).$$But wait--we said $x = 2 heta$ ! Plugging this back in allows us to conclude that$$frac12 sin(4 heta) = sin(2 heta)cos(2 heta).$$


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Note:eginequationsin(2x) = 2sin(x)cos(x) \endequation

Therefore we have $fracsin(4x)2 = fracsin(2(2x))2= frac2sin(2x)cos(2x)2=sin(2x)cos(2x)$

Does this clarify?


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If you are familiar with the double angle identity:

$sin(2x) = 2sin(x)cos(x)$ or written another way

$1 over 2 sin(2x) = sin(x)cos(x)$

Let $x = 2 heta$ & you"ll get:

$1 over 2 sin(4 heta) = sin(2 heta)cos(2 heta)$


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